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3 years ago

Solve the equation using the quadratic formula : x^2-10x+25=18

1 answer:

8 0

$x^2-10x+25=18 \\ \\x^2-10x+25-18 = 0\\ \\ x^2-10x+ 7 =0 \\ \\a=1 , b = -10 , c= 7 \\ \\ \Delta = b^{2}-4ac = (-10)^{2}-4*1*7=100-28 = 72 \\ \\\sqrt{\Delta }=\sqrt{72}= \sqrt{2*36} =\sqrt{36}*\sqrt{2} =6\sqrt{2}$

$x_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{10- 6\sqrt{2}}{2}=\frac{2(5-3\sqrt{2})}{2}= 5-3\sqrt{2}\\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a} =\frac{10+ 6\sqrt{2}}{2}=\frac{2(5+3\sqrt{2})}{2}= 5+3\sqrt{2}$

$Answer : \ x= 5-3\sqrt{2} \ \ or \ \ x = 5+3\sqrt{2}$

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you secure a mortgage to buy a house with a loan of $140,000 at 8.5% for 20 years. answer the following questions about that loa

Yuri [45]

Answer:

monthly payment $1214.95
1st month's interest $991.67
balance after 1st payment $139,776.72
2nd month's interest $990.09
balance after 2nd payment $139,551.86

Step-by-step explanation:

The monthly interest rate is ...

$\dfrac{8\%}{12}=0.00708\overline{3}$

a) The monthly payment is given by the amortization formula:

A = Pr/(1 -(1+r)^-n)

where r is the monthly interest rate on a loan of amount P for n months.

A = $140,000(0.0070833)/(1 -(1.0070833^-240)) = $1214.95

The monthly payment is $1214.95.

b) The amount to interest is the product of the remaining principal and the monthly interest rate.

first month's interest = $140,000·0.0070833 = $991.67

c) The balance after the first payment is ...

new balance = $140,000 +991.67 -1214.95 = $139,776.72

d) The amount to interest for the second payment is computed the same way:

second month's interest = $139,776.72·0.00708333 = $990.09

e) The balance after the second payment is computed the same way:

new balance = $139,776.72 +990.09 -1214.95 = $139,551.86

8 0

3 years ago

Use z scores to compare the given values.

Effectus [21]

Answer:

The answer is given below

Step-by-step explanation:

Z score is used to measure by how many standard deviations the raw score is above or below the mean. It is given by the formula:

$z=\frac{x-\mu}{\sigma}\\ \\Where\ \mu=mean, x=raw\ score, \sigma=standard\ deviation$

For the male:

μ = 3259.8 g, σ = 869.7 g

Hence for a male who weighs 1600 g, the z score is:

$z=\frac{x-\mu}{\sigma} =\frac{1600-3259.8}{869.7}=-1.91$

For the female:

μ = 3000.9 g, σ = 550.8 g

Hence for a female who weighs 1600 g, the z score is:

$z=\frac{x-\mu}{\sigma} =\frac{1600-3000.9}{550.8}=-2.54$

Since the z score for the male is z= -1.91 and the z score for the female is z= -2.54 the male has the weight that is more extreme.

6 0

2 years ago

What is the slope of the following graph? (write your answer as a fraction unless divided by 1. Do not use spaces in your answer

alexira [117]

The slope of the graph is 4/5

<h3>How to determine the slope</h3>

Using the formula

Slope = Δ y-axis / Δ x- axis

y2 = -4

y1 = 0

x2 = -5

x1 = 0

Slope = $\frac{y2 - y1}{x2 - x1}$

Substitute the values into the formula

<em>Slope = </em> $\frac{-4 - 0}{- 5 - 0}$

Slope = $\frac{-4}{-5}$

Slope = $\frac{4}{5}$

Therefore, the slope of the graph is 4/5

Learn more about slope here:

brainly.com/question/3493733

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5 0

2 years ago

Divide 14 by v. Then, subtract 7.

Nastasia [14]

Answer:

14-7v

--------

Step-by-step explanation:

i dont understand what else the answer could be, due to the fact that not enough information if provided to result in a solid sum.

3 0

3 years ago

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an

maxonik [38]

Answer:

5.0 ft-lbf

Step-by-step explanation:

The force is

$F = \dfrac{9}{6^x}$

This force is not a constant force. For a non-constant force, the work done, <em>W</em>, is

$W = \int\limits^{x_2}_{x_1} {F(x)} \, dx$

with $x_1$ and $x_2$ the initial and final displacements respectively.

From the question, $x_1 =0$ and $x_2 = 12$ .

Then

$W = \int\limits^{12}_0 {\dfrac{9}{6^x}} \, dx$

Evaluating the indefinite integral,

$\int\limits \dfrac{9}{6^x} \, dx =9 \int\limits\!\left(\frac{1}{6}\right)^x \, dx$

From the rules of integration,

$\int\limits a^x\, dx = \dfrac{a^x}{\ln a}$

$9 \int\limits \left(\frac{1}{6}\right)^x \, dx = 9\times\dfrac{(1/6)^x}{\ln(1/6)} = -5.0229\left(\dfrac{1}{6}\right)^x$

Returning the limits,

$\left.-5.0229\left(\dfrac{1}{6}\right)^x\right|^{12}_0 = -5.0229(0.1667^{12} - 0.1667^0) = 5.0229 \approx 5.0 \text{ ft-lbf}$

4 0

3 years ago